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An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colours is
[AIEEE 2010]
- a)1/3
- b)2/7
- c)1/21
- d)2/23
Correct answer is option 'B'. Can you explain this answer?
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Problem:
An urn contains nine balls of which three are red, four are blue, and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colors is?
Solution:
To find the probability that the three balls have different colors, we need to find the total number of favorable outcomes and the total number of possible outcomes.
Step 1: Total number of possible outcomes
When three balls are drawn without replacement, the total number of possible outcomes can be found using combinations. We need to choose 3 balls out of the 9 available balls.
Number of possible outcomes = C(9, 3) = 84
Step 2: Total number of favorable outcomes
For the three balls to have different colors, we can consider the following cases:
Case 1: Red, Blue, Green
- We can choose 1 red ball from the 3 available red balls.
- We can choose 1 blue ball from the 4 available blue balls.
- We can choose 1 green ball from the 2 available green balls.
- Number of favorable outcomes = C(3, 1) * C(4, 1) * C(2, 1) = 3 * 4 * 2 = 24
Case 2: Blue, Red, Green
- We can choose 1 blue ball from the 4 available blue balls.
- We can choose 1 red ball from the 3 available red balls.
- We can choose 1 green ball from the 2 available green balls.
- Number of favorable outcomes = C(4, 1) * C(3, 1) * C(2, 1) = 4 * 3 * 2 = 24
Case 3: Blue, Green, Red
- We can choose 1 blue ball from the 4 available blue balls.
- We can choose 1 green ball from the 2 available green balls.
- We can choose 1 red ball from the 3 available red balls.
- Number of favorable outcomes = C(4, 1) * C(2, 1) * C(3, 1) = 4 * 2 * 3 = 24
Case 4: Green, Blue, Red
- We can choose 1 green ball from the 2 available green balls.
- We can choose 1 blue ball from the 4 available blue balls.
- We can choose 1 red ball from the 3 available red balls.
- Number of favorable outcomes = C(2, 1) * C(4, 1) * C(3, 1) = 2 * 4 * 3 = 24
Total number of favorable outcomes = 24 + 24 + 24 + 24 = 96
Step 3: Calculating probability
Probability = Number of favorable outcomes / Total number of possible outcomes
Probability = 96 / 84
Probability = 2/7
Therefore, the probability that the three balls have different colors is 2/7.
An urn contains nine balls of which three are red, four are blue, and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colors is?
Solution:
To find the probability that the three balls have different colors, we need to find the total number of favorable outcomes and the total number of possible outcomes.
Step 1: Total number of possible outcomes
When three balls are drawn without replacement, the total number of possible outcomes can be found using combinations. We need to choose 3 balls out of the 9 available balls.
Number of possible outcomes = C(9, 3) = 84
Step 2: Total number of favorable outcomes
For the three balls to have different colors, we can consider the following cases:
Case 1: Red, Blue, Green
- We can choose 1 red ball from the 3 available red balls.
- We can choose 1 blue ball from the 4 available blue balls.
- We can choose 1 green ball from the 2 available green balls.
- Number of favorable outcomes = C(3, 1) * C(4, 1) * C(2, 1) = 3 * 4 * 2 = 24
Case 2: Blue, Red, Green
- We can choose 1 blue ball from the 4 available blue balls.
- We can choose 1 red ball from the 3 available red balls.
- We can choose 1 green ball from the 2 available green balls.
- Number of favorable outcomes = C(4, 1) * C(3, 1) * C(2, 1) = 4 * 3 * 2 = 24
Case 3: Blue, Green, Red
- We can choose 1 blue ball from the 4 available blue balls.
- We can choose 1 green ball from the 2 available green balls.
- We can choose 1 red ball from the 3 available red balls.
- Number of favorable outcomes = C(4, 1) * C(2, 1) * C(3, 1) = 4 * 2 * 3 = 24
Case 4: Green, Blue, Red
- We can choose 1 green ball from the 2 available green balls.
- We can choose 1 blue ball from the 4 available blue balls.
- We can choose 1 red ball from the 3 available red balls.
- Number of favorable outcomes = C(2, 1) * C(4, 1) * C(3, 1) = 2 * 4 * 3 = 24
Total number of favorable outcomes = 24 + 24 + 24 + 24 = 96
Step 3: Calculating probability
Probability = Number of favorable outcomes / Total number of possible outcomes
Probability = 96 / 84
Probability = 2/7
Therefore, the probability that the three balls have different colors is 2/7.
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An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colours is[AIEEE 2010]a)1/3b)2/7c)1/21d)2/23Correct answer is option 'B'. Can you explain this answer?
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